Chapter EH, Problem 33R

Solution

Interpretation: The electron configuration of tin needs to be determined.

Concept Introduction: The electronic configuration represents the arrangement of electrons in atomic orbitals. Here, s, p, d, and f are sublevels in which electrons are distributed. The maximum number of electrons in s, p, d, and f orbitals are 2, 6, 10, and 14.

  $\left[\text{Kr}\right]4d^{10}5s^{2}5p^2$

The given element is tin. It belongs to group 14. The atomic number of tin is 50.

The electron configuration can be represented as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^2$

Here, the 4s subshell is filled before 3d .

This can be rewritten as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^2$

Now, in shorthand notation, the nearest noble gas is mentioned and electron configuration can be written including it. Here, the nearest noble gas is Kr with atomic number 36.

Thus, the shorthand notation for the above electron configuration will be:

  $\left[\text{Kr}\right]4d^{10}5s^{2}5p^2$

Therefore, the electronic configuration of tin is $\left[\text{Kr}\right]4d^{10}5s^{2}5p^2$ .